We note first that division by zero is not allowed and that for this reason \(3x-9\) may not be equal to zero and that therefore \(x=3\) is not a solution.

We now distinguish two cases, namely \(3x-9>0\) and \(3x-9<0\).
In both cases we multiply the inequality on both sides by \(3x-9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(3x-9>0\), i.e. \(x> 3\). Then we get \(1<3x-9\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-3x<-10\).
Then, dvision by the coefficient of \(x\)gives \(x > {{10}\over{3}}\).
So we have the following system of inequalities: \(x> 3\,\wedge\; x > {{10}\over{3}}\)
and this simplifies to \(x\gt{{10}\over{3}}\).

Suppose \(3x-9<0\), i.e. \(x< 3\). Then we get \(1>3x-9\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-3x>-10\).
Then, division by the coefficient of \(x\) gives \(x < {{10}\over{3}}\).
So we have the following system of inequalities: \(x< 3\,\wedge\; x < {{10}\over{3}}\)
and this simplifies to \(x\lt 3\).

The solution of the original inequality is \(x\lt 3\;\vee\;x\gt{{10}\over{3}}\).